Answer:
4.15 m/s
Explanation:
Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.
i.e.
acceleration = a = - 0.1 m/s²
Distance covered = S = 6m
Velocity after covering 6 meters = Final velocity = 
= 4 m/s
We need to find the initial speed, which will be the same as the magnitude of initial velocity.
Initial velocity = 
= ?
3rd equation of motion relates the acceleration, distance, final velocity and initial velocity as:
Using the known values in the formula, we get:
Thus, the initial speed of the ball was 4.15 m/s
Answer:
The answer is the 1st one
What is the question asking?
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
= + 2as
= + 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
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